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MATH A305 Introduction to Geometries
Final Exam Key
Instructions
1. Do NOT write your answers on these sheets. Nothing written on the test papers will be graded.
2. Please begin each section of questions on a new sheet of paper.
3. Do not write problems side by side.
4. Do not staple test papers.
5. Limited credit will be given for incomplete or incorrect justification.
Questions
1. Tesselate
(a) Euclidean
i. Note that rectangles can be used to tesselate the plane because four rectangles arranged around a
point have four right angles adding up to 2π. Similarly equilateral triangles (6 around a point for
6 · π/3 = 2π) and regular hexagons (3 around a point for 3 · 2π/3 = 2π) can be used to tesselate.
Any triangle can be used to tesselate by rotating the triangle around the midpoint of one side by π
then translating copies of this pair of triangles. Reference Figure 1. Prove that this works.
Note that the rotation about the midpoint of a side places opposite angles next to each other. By
next translating the resulting quadrilateral along both other sides, two copies of each angle are placed
next to each other.
Because the angle sum of a triangle is π this gives an angle sum of 2π so the corners all add up as
needed.
ii. For signature *632 the minimum piece is outlined by the lines of reflection. Determine how the
minimum shape is connected to the isometries for signatures: 2222, o, *X.
As shown in Figure 2 to 4 the minimum pieces are as follows.
2222: The region is defined by the four lines through two copies of two points of rotations by π. Note
the other two rotations are also on the lines. This forms a quadrilateral.
o: The region is defined by two copies of each of the vectors. These form a parallelogram (quadrilateral).
*X: The region is defined by two lines of reflection and a height determined by the translation vector
from the glide reflection. This is another quadrilateral.
(b) Hyperbolic
In hyperbolic geometry every triangle has angle sum less than π. The difference D(4ABC) = π −
m6 ABC + m6 BCA + m6 CAB is called the defect. The defect is additive, so the defect of a rectangle is
the defect of any two triangles into which it can be broken.
i. It is possible to tesselate with a regular pentagon. Explain why.
The exterior angles of a Euclidean pentagon are measure 2(π − 2π/5)/2 = 3π/5. In a hyperbolic plane
the angles are 3π/5 − d (defect). Pentagons will tesselate for any defect such that 3π/5 − d = 2π/k.
The smallest is k = 4 (because smaller k give a negative solution for defect).
ii. How many distinct, regular pentagons can tesselate?
There are infinitely many because the defect approaches 3π/5 allowing many solutions to the equation
above.
iii. How many pentagons can be arranged around a point? Note in Euclidean this would be four for
rectangles as noted above.
As shown above the minimum is 4. As the defect increases the number of pentagons increases.
1
Final Exam
2
2. Symmetry
(a) Using a matrix transformation prove that the image in Figure 5 has four fold rotational symmetry about
the origin. For this step prove just that the vertices are symmetrical.
The rotation is by π/2 about the origin.

Rπ/2

3 0
Rπ/2  0 3
1 1

 
cos(π/2) − sin(π/2) 0
x
0
cos(π/2) 0   y  =  1
=  sin(π/2)
0
0 1
1
0
√
√
−3
0 √ 2 −√ 2
0 −3
2
2
1
1
1
1
√
−√ 2
− 2
1
√  
0
√2
− 2 = 3
1
1

 

−1 0
x
−y
0 0  y  =  x 
0 1
1
1
√
√
√ 
√
−3
0 3 −√2 −√2
√2 √ 2
0 −3 0
2 − 2 − 2
2 
1
1 1
1
1
1
1
Note that each point permuted to the next. Repeating four times would cycle back to the beginning as a
result.
(b) Prove that one of the lines is symmetrical using the matrix.
√
√
√ √
One of the lines is from ( 2, 2) to (0, 3). The equation is [0, 3, 1] − [− 2, 3 − 2, 0]t. Note
√  
√



0 −1 0
−3 + (3 − √2)t
√ 2t
 1
0 0   3 − (3 − 2)t  = 
2t 
0
0 1
1
1
This is another one of the lines.
Final Exam
3
3. Transformations
(a) Find the parameters of a similarity for each of the similarities provided.
~ 00 . Then dilate the triangle by 2 from C. Then
See Figure 6. The steps were to translate the triangle by CC
connect corresponding points of this newly dilated triangle with the original to find the point of dilation.
The reflection line is through C orthogonal to BB 00 .
See Figure 7. The steps were to copy length of AB to A00 B 00 . Next construct the segments AA00 and BE
where E came from the copy above. The perpendicular bisectors of these segments provide the center of
rotation. The center of dilation is from A00 .
(b) For one similarity find a reversed (isometry and dilation) similarity.
See Figure 8. Extend AC and A00 C 00 . Rotate the triangle about A by the angle between these lines.
Connect corresponding vertices. The intersection of these lines give the point of dilation.
(c) Prove that a translation followed by a reflection orthogonal to the vector of translation is just a reflection.
~ 0 . The reflection is over l. By definition of reflection B 0 B 00 ⊥ l.
See Figure 9. The translation is along BB
~ 0 , B − B 0 − B””.
Because the line of reflection l is orthogonal to the vector of translation BB
Construct line m as the perpendicular bisector of BB 00 . Thus B and B 0 are equidistant from m on a line
orthogonal to m.
The same is true of the other points.
(d) (Bonus) Prove that given the pre-image and image of a glide reflection, that a representation is a translation
from one point to its corresponding image point followed by a reflection about the perpendicular bisector
of the segments connecting either pair of other corresponding points.
(e) Give the matrix for a glide reflection with translation vector (2, 7) and reflection over the line y = (x − 3).
translate

1 0
 0 1
0 0
back
x and 
y
 swap

3
0 1 0
0   1 0 0 
1
0 0 1
translate to origin


1 0 −3
 0 1
0 
0 0
1
glide

1 0
 0 1
0 0

2
7 
1
Final Exam
4
A'
B C'
A
B C'
β
A
γ
C B'
β
α
α
α A' γ
γ β
α
β
C B'
Figure 1: Tesselating with Random Triangle
Final Exam
5
Figure 2: 2222
Figure 3: o
Final Exam
6
Figure 4: *X
(0,3)
(-3,0)
(- 2, 2)
( 2, 2)
(- 2,- 2)
( 2,- 2)
(0,-3)
Figure 5: Analytic Symmetry
(3,0)
Final Exam
7
B''
H
B'1
G
A
B
A'1
A''
C''1
IC'
C
Figure 6: Dilation 1
Final Exam
8
C''
C'1
B''
B
B'
E1
C
A''1
A'
A
Figure 7: Dilation 2
Final Exam
9
C''
C'1
B''
B
B'1
C
A''
A1
A'
Figure 8: Dilation 2 Reverse
B
A
B'
C
A'
m
l
M
R
C'
Figure 9: Just a Reflection
B''
C''
A''